Question

条件变量的典型用法如下所示（参见下面的代码）：http://en.cppreference.com/w/cpp/thread/condition_variable。

但是，似乎主线程可能在工作线程调用notify_one之前调用wait，这会导致死锁。我错了吗？如果没有，通常的解决方法是什么？

#include <iostream>
#include <string>
#include <thread>
#include <mutex>
#include <condition_variable>

std::mutex m;
std::condition_variable cv;
std::string data;
bool ready = false;
bool processed = false;

void worker_thread()
{
    // Wait until main() sends data
    std::unique_lock<std::mutex> lk(m);
    cv.wait(lk, []{return ready;});

    // after the wait, we own the lock.
    std::cout << "Worker thread is processing data\n";
    data += " after processing";

    // Send data back to main()
    processed = true;
    std::cout << "Worker thread signals data processing completed\n";

    // Manual unlocking is done before notifying, to avoid waking up
    // the waiting thread only to block again (see notify_one for details)
    lk.unlock();
    cv.notify_one();
}

int main()
{
    std::thread worker(worker_thread);

    data = "Example data";
    // send data to the worker thread
    {
        std::lock_guard<std::mutex> lk(m);
        ready = true;
        std::cout << "main() signals data ready for processing\n";
    }
    cv.notify_one();

    // wait for the worker
    {
        std::unique_lock<std::mutex> lk(m);
        cv.wait(lk, []{return processed;});
    }
    std::cout << "Back in main(), data = " << data << '\n';

    worker.join();
}

Answer 1

请注意使用条件的definition of wait（您应该使用的唯一等待时间）：

while (!pred()) {
    wait(lock);
}

如果已经触发了notify，则表示条件已经为真（在信令线程中的notify_one之前已经排序）。因此，当接收器使用互斥量并查看pred（）时，它将成立并且将继续。

如何确保在notify_one之前调用wait_for

1 个答案: